Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only th... | You can substitute $u=x+y$ and $v=xy$. Then
$$x^2+xy+y^2=(x+y)^2-xy=u^2-v $$
and
$$x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=(u^2-2v)^2-v^2 $$
You obtain the system
$$\begin{cases}
u^4-4u^2v+3v^2=21\\ u^2-v=3
\end{cases}$$
From the second equation you get $v=u^2-3$. Substitute that in the first equation, and you'll immediately... | {
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"url": "https://math.stackexchange.com/questions/3500677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with inequality problem
Given $a$ , $b$ , $c \ge 0$ show that
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$
I tried using Titu's lemma on it, resulting in
$$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2 + 3(... | Hint.
As the inequality is homogeneous, making the substitutions $b=\lambda a, c=\mu a$ we have
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} = \frac{\lambda ^2 \mu +\lambda ^2+\lambda \mu ^2+\lambda +\mu ^2+\mu }{(\lambda +1) (\mu +1) (\lambda +\mu )} = f(\lambda,\mu)
$$
with stationary po... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving logarithm leaving in terms of $p$ and $q$ I would like to check the steps if Part a) is done correctly.
For Part b), how do I continue from below? I seem to stuck for $\log_{10}(5)$…
Here is the problem:
Given that $p = \log_{10} 2$ and $q = \log_{10} 7$, express the following in terms of $p$ and $q$.
a) $\log... | Hint: Write $5 = 10/2$. ${}{}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Element of order $7$ in $GL(4,2)$ Find an element of order $7$ in $GL(4,2)$, the group of all invertible $4 \times 4$ matrices with entries in $\mathbb{F}_2$.
I'd like a more a constructive way to find the required element in $GL(4,2)$ in lieu of explicity going through elements of $GL(4,2)$ and computing their orders... | If $\lambda\in\mathbb{F}_{2^3}$ is such that $\lambda^3=\lambda+1 $ then $\lambda^6=\lambda^2+1 $ and $\lambda^7 = \lambda^3+\lambda = 1 $, so the companion matrix of $x^3-x-1$ (as an element of $\text{GL}(3,\mathbb{F}_2)$) is a $3\times 3$ matrix with order $7$, which can be easily completed to a $4\times 4$ matrix wi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Correlation between two matrices what is the correlation/relationship between two matrices
A = \begin{bmatrix}4&0&0&0\\0&3&0&0\\0&0&2&0\end{bmatrix}
B = \begin{bmatrix}1/4&0&0\\0&1/3&0\\0&0&1/2\\0&0&0\end{bmatrix}
where A*B = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
A cannot be an inverse of B because it's not s... | You may be reaching for the Moore-Penrose pseudoinverse, usually denoted as $A^+$, which is also defined for rectangular and singular matrices.
In this case:
$$B^+ = \begin{bmatrix}1/4&0&0\\0&1/3&0\\0&0&1/2\\0&0&0\end{bmatrix}^+
= \begin{bmatrix}4&0&0&0\\0&3&0&0\\0&0&2&0\end{bmatrix}$$
And:
$$B^+\cdot B=\begin{bmatrix}... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $y' + y^2 = \frac{1}{x^2}$ by introducing $z = xy$ as a new function. Question:
Solve the equation:
$$y' + y^2 = \frac{1}{x^2},~~~~~~~~~x > 0$$
by introducing $z = xy$
Attempted answer:
$z = xy \Rightarrow y = \frac{z}{x}$
Taking the derivative of $y$ with respect to $x$ using he product rule:
$$y' = \frac{z'}{x}... | Because of the division that must be different from zero as GSofer pointed in his good answer but you can also deduce that solution by other methods:
$$y' + y^2 = \frac{1}{x^2}$$
By inspection $y_p=\frac A x$
$$-\frac A {x^2}+\frac {A^2}{x^2}=\frac 1 {x^2}$$
$$ \implies A^2-A-1=0$$
$$\implies A=\frac 12 \pm\frac {\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?
Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used ... | Assume that a $5$ is first seen on roll $n$.
$5$ is the lowest seen with $n-1$ $6$s then one $5$.
$4$ is the lowest seen with $n-1$ $4$s and $6$s, but not all $6$s then one $5$.
$3$ is the lowest seen with $n-1$ $3$s, $4$s, and $6$s, but not all $4$s and $6$s then one $5$.
$2$ is the lowest seen with $n-1$ $2$s, $3$s, ... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $2^m=7n^2+1$ Solve $2^m=7n^2+1$ with $(m,n)\in \mathbb{N}^2$
Here is what I did:
First try, I have seen first that the obvious solutions are $n=1$ and $m=3$ , and $n=3$ and $m=6$, then I proved by simple congruences that $m$ must be divisible by $3$ so $m=3k$, If we add $27$ to the equation we will have $2^{3k}+3... | HINT
$2^m\equiv 1$ mod $7$ and so $m=3k$. For $n>0$, we now have
$$2^k-1=au^2, 2^{2k}+2^k+1=bv^2$$
where either $\{a,b\}=\{1,7\}$ or $\{a,b\}=\{3,21\}.$
Each of the four possibilities gives an elliptic curve $$bv^2=3+3au^2+a^2u^4.$$
Of these, the case $a=7,b=1$ is impossible modulo $7$.
| {
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"url": "https://math.stackexchange.com/questions/3506091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Easy double sum trouble : $\sum_{1 \leq i,j \leq n} (i+j)^2$ There's a mistake somewhere but I really can't see where.
$\sum_{1 \leq i,j \leq n} (i+j)^2 = \sum_{i=1}^n \sum_{j=1}^n \left( i^2+2ij+j^2\right) = \sum_{i=1}^n \left( \sum_{j=1}^n i^2 + 2i\sum_{j=1}^n j + \sum_{j=1}^n j^2\right) $
$= \sum_{i=1}^n\left( ni^2... | The last number is $(3+3)^2$ which is $36$.
And here is analytic result $$\frac{1}{6} \left(7 n^4+12 n^3+5 n^2\right)$$
| {
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"url": "https://math.stackexchange.com/questions/3507225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.
$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$
$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\... | Hint:
If $$f(x)=\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$$ $\forall$ $\vert x\vert \lt 1$. Then what is $$\frac{f'\left(\frac{1}{2}\right)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to solve following differential equation $y = \frac{xy'}{2} + \frac{y'^2}{x^2}$? Given the following differential equation:
$$y = \frac{xy'}{2} + \frac{y'^2}{x^2}$$
I tried to solve it downgrading and as a quadratic equation for unknown $y'$ but it did not bring any results
UPD Possible solution:
$$y = \frac{xy'}{2... | Hint. If we differentiate the equation
$$
y = \frac{xy'}{2} + \frac{y'^2}{x^2}
$$
we obtain
$$
y'=\frac{y'}{2}+\frac{xy''}{2}+\frac{2y'y''}{x^2}-\frac{2y'^2}{x^3}
$$
or
$$
-\frac{y'}{2}+\frac{xy''}{2}+\frac{2y'y''}{x^2}-\frac{2y'^2}{x^3}=0.
$$
or
$$
(xy''-y')\left(\frac{1}{2}+\frac{2y'}{x^3}\right)=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $mx+3|x+4|-2=0$ has no solutions, solve for $m$
If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?
(A)5
(B)$-\frac{1}{2}$
(C)-3
(D)-6
(E)$\frac{10}{3}$
my attempt:
$$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$
because the equation has no solutions, therefor... | The hint.
For $x\geq-4$ we obtain:
$$mx+3x+12-2=0,$$ which gives a value $m=-3$.
For $x\leq-4$ we obtain $$mx-3x-12-2=0,$$ which gives a value $m=3.$
Now, check that for $m=3$ our equation has root, while for $m=-3$ our equation has no roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Linear differential equations, integrating factor Solve the following differential equation:
$$ dr+(2r \cot\theta+\sin 2\theta)d\theta=0$$
I have tried like this:
$$ \frac{dr}{d\theta}+2r\cot\theta=-\sin{2\theta}$$
\begin{align}
I.F. &=e^{\int{2\cot\theta d\theta}}\\
& =e^{-2\log\sin\theta}\\
& =\frac 1{\sin^2\theta}\\... | $$dr+(2r \cot\theta+\sin 2\theta)d\theta=0$$
$$\sin \theta dr+(2r \cos\theta+2\sin^2 \theta\cos\theta )d\theta=0$$
Multiply by $2\sin \theta$:
$$2\sin^2 \theta dr+(4r \cos\theta\sin \theta+4\sin^3 \theta\cos\theta )d\theta=0$$
$$2\sin^2 \theta dr+2r d\sin^2 \theta+ d\sin^4 \theta=0$$
$$2(\sin^2 \theta dr+r d\sin^2 \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Inverse of $I + A$ I am trying to solve the following exercise in Artin, without breaking into cases for even and odd $k$.
A square matrix $A$ is called nilpotent if $A^k = 0$ for some $k > 0$. Prove that if $A$ is nilpotent, then $I + A$ is invertible.
Here's my attempt.
I claim that the inverse of $I + A$ is
$$... | You could use
$$
\sum\limits_{n=0}^{k-1} (-1)^n (A^n + A^{n+1}) =\sum\limits_{n=0}^{k-1} (-1)^n A^n +\sum\limits_{n=0}^{k-1} (-1)^n A^{n+1} = \sum\limits_{n=0}^{k-1} (-1)^n A^n -\sum\limits_{n=0}^{k-1} (-1)^{n+1} A^{n+1} =\sum\limits_{n=0}^{k-1} (-1)^n A^n -\sum\limits_{n=1}^{k} (-1)^n A^n = I - (-1)^kA^k
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find: $\lim_{x\to 0}\frac{\ln(e+x^2)-\cos x}{e^{x^2}-\cos x}$
Evaluate without L'Hospital's rule: $\displaystyle\lim_{x\to 0}\frac{\ln(e+x^2)-\cos x}{e^{x^2}-\cos x}$
My attempt:
$$e^{x^2}-\cos x=e^{x^2}-1+1-\cos x=x^2\cdot\left(\frac{e^{x^2}-1}{x^2}+\frac{1-\cos x}{x^2}\right)$$
$$\frac{\ln(e+x^2)-\cos x}{x^... | $$L=\lim_{x \rightarrow 0} \frac{\ln (e+x^2)-\cos x}{e^{x^2}-\cos x}$$ Let us use
$$e^{x^2}=1+x^2+O(x^4), \cos x= 1+ x^2/2+O(x^4), \ln(1+z)=z-z^2/2+O(z^4),$$
then $$L=\lim_{x \rightarrow 0}\frac{1+x^2/e-1+x^2/2+O(x^4)}{3x^2/2+O(x^4)}=\frac{2+e}{3e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A generalization of the (in)famous IMO 1988 problem 6: If $\frac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square. This question is motivated by the famous IMO $1988$ problem $6$. Is the following true?
Let $a,b$ be positive integers and $c \ge 0$ be a non-negative integer. If $\dfrac{a^2 + b^2 - a... | We will adapt the proof for $c=0$. So let us consider $c\ge 0$ to be fixed.
Assume we have a solution $(a,b)$ of the problem, so that the number
$$
k =\frac{a^2+b^2-abc}{ab+1}\tag{$1$}
$$
is an integer, $k\in \Bbb Z$, and it is positive, $k>0$.
We can and do assume that $a>b\ge 0$. (The case $a=b$ is easily eliminated... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to multiply $\sqrt{5x^2+2x+1}$ with $\frac{1}{x}$? I believe the answer says, $\sqrt{5+\frac{2}{x}+\frac{1}{x^2}}$, but I don't see how they obtained it. I'm quite rusty with radicals.
| If $x>0$ then $x=\sqrt{x^2}$. So if $x>0$, then
$$\begin{align*}
\frac{1}{x}\cdot\sqrt{5x^2+2x+1} &= \frac{\sqrt{5x^2+2x+1}}{x} \\
&= \frac{\sqrt{5x^2+2x+1}}{\sqrt{x^2}} \\
&= \sqrt{\frac{5x^2+2x+1}{x^2}} \\
&= \sqrt{\frac{5x^2}{x^2}+\frac{2x}{x^2}+\frac{1}{x^2}} \\
&= \sqrt{5+\frac{2}{x}+\frac{1}{x^2}} \\
\end{align*}... | {
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"timestamp": "2023-03-29T00:00:00",
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Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1... | Algebra students learn that the quadratic formula says
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
When is this true? It's helpful to remember that the quadratic formula comes from the following statement, which is true for any complex numbers $a$, $b$, $c$, and $x$ (as long as $a\ne0$):
$$\text{If }ax^2+bx+c=0,\quad\text{th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Prove that $A^2=a_1^2+a_2^2-a_3^2-a_4^2$ for all integers $A$.
Prove that:
For every $A\in\mathbb Z$, there exist infinitely many $\{a_1,a_2,a_3,a_4\}\subset\mathbb Z$ given $a_m\neq a_n $ such that $$A^2=a_1^2+a_2^2-a_3^2-a_4^2$$
After many hours, I found a general formula satisfying the statement for all $A$ and ... | Given any non-negative integer $A$, let $a_1=A+2n+1$ for any of $n=1,2,3,\cdots$.
For convenience let $b=a_1^2-A^2 = (A+2n+1)^2-A^2= (4n+2)A+4n^2+4n+1$. Thus $b$ is odd and $\geq9$.
Now let $a_2=(b-1)/2$ and $a_3=(b+1)/2$, implying that $a_2^2-a_3^2=-b$, and let $a_4=0$. Then:
$$a_1^2+a_2^2-a_3^2-a_4^2=a_1^2+(-b)-0=b+... | {
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"timestamp": "2023-03-29T00:00:00",
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If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy} {ab}\cos(\theta)$ If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, prove
that $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos(\theta)=\sin^2(\theta)$
My trial:Let:
$\cos^{-1}(\frac{x}{a})=\alpha$, ... | Hint
Using $\cos(A+B)$ formula
$$\cos\theta-\dfrac{xy}{ab}=-\sqrt{\left(1-\dfrac{x^2}{a^2}\right)(\cdots)}$$
Take square in both sides
Alternatively
Let $\cos^{-1}\dfrac xa=A,\cos A=?$ etc.
$$\cos\theta=\cos(A+B)$$
Rearrange and square both sides
$$(\cos\theta-\cos A\cos B)^2=(-\sin A\sin B)^2=(1-\cos^2A)(1-\cos^2B)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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First order PDE problem $yu_{x} - xu_{y} = x^2$ I am a beginner to PDE, trying to solve $$yu_{x} - xu_{y} = x^2$$ using characteristic line, first I have \begin{align*}
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2}
\end{align*}
Then I get the characteristic line is given by $$C = \frac{1}{2}y^2 - \frac{1}{2}x^2$$
N... | $$yu_x-xu_y=x^2$$
Charpit-Lagrange system of characteristic ODEs:
$$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{x^2}$$
A first characteristic equation comes from $\frac{dx}{y}=\frac{dy}{-x}$
$$x^2+y^2=c_1$$
A second characteristic equation comes from $\frac{dx}{\sqrt{c_1-x^2}}=\frac{du}{x^2}$
$du=\frac{x^2dx}{\sqrt{c_1-x^2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$
Given $\triangle ABC$, (using the main parameters and notation), prove that $$ \frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2 \cdot \left(\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \ta... | In the standard notation we need to prove that:
$$\sum_{cyc}\frac{a}{\dfrac{2S}{a}}\geq2\sum_{cyc}\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{1+\frac{b^2+c^2-a^2}{2bc}}}$$ or
$$\sum_{cyc}a^2\geq4S\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{(b+c-a)(a+b+c)}}$$ or
$$\sum_{cyc}a^2\geq\sum_{cyc}(a^2-(b-c)^2),$$ which is obvious.
| {
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"url": "https://math.stackexchange.com/questions/3524570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-... | Wow. There are many good solutions here.
@James Warthington, I'd like to add another approach but using a more methodical application of the Chain Rule.
$y=(3+x)^{-1}(3-x^2)^ \frac{1}{2}$
First variable substitution, let $\mu=3+x$, and note(for later) that $\frac{d}{dx}\mu=1$
Second variable substitution, let $\omega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Evaluate $\int_{0}^{\infty} \frac{x^2+x+1}{x^6+x^4+1} dx$ $$\int_{0}^{\infty} \frac{x^2+x+1}{x^6+x^4+1} dx $$
Wolfram says $1.80276\ldots $
Calculating this seems complicated to me because the residues are pretty hard to find and I have tried the infinite residue method but it didn't work out.
| Without complex analysis.
From a formal point of view, there is closed form solution.
Let $(a,b,c)$ to be the roots of $y^3+y^2+1=0$; one is real and negative (say $a$) and the other two $(b,c)$ are complex conjugate.
$$\frac{x^2+x+1}{x^6+x^4+1}=\frac{x^2+x+1}{(x^2-a)(x^2-b)(x^2-c)}$$ Now, using partial fraction decomp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Difference between sinc and $\cos$ can be expressed using Bessel function $J_{3/2}$ I've tried to prove this property of Bessel function but I don't seem to be going anywhere
$$\sqrt{\frac 12 \pi x} J_\frac 32 (x) = \cfrac{\sin x}{x} - \cos x$$
I have tried substituting $\frac 32$ for $J_n (x)$ and then manipulating wi... | Using the series expansion of $J_{3/2}(x)$ and the Legendre duplication formula for the gamma function, we find
$$
\sqrt {\frac{{\pi x}}{2}} J_{3/2} (x) = \sqrt {\frac{{\pi x}}{2}} \left( {\tfrac{1}{2}x} \right)^{3/2} \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{\left( {\frac{1}{4}x^2 } \right)^n }}{{n!\Gamma \left( {n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
integration of. $\int \frac{x}{x^3-3x+2}$ I am trying to integrate :
$\Large \int \frac{x}{x^3-3x+2}dx$
I decomposed the fraction and got :
$\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions:
$ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
W... | You don't have all the components in the decomposition. It should be
$$\frac {x}{(x-1)^2(x+2)} = \frac {Ax+C}{(x-1)^2} + \frac {B}{x+2}
=\frac {\frac29x+\frac19}{(x-1)^2} - \frac {\frac29}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $\frac{1}{a+3}+\frac{1}{b+4}+\frac{1}{c+5}=\frac{7}{12}$ for positive integer $a$, $b$, $c$, then find $\frac{a}{a+3}+\frac{b}{b+4}+\frac{c}{c+5}$
Given $a, b, c$ are positive integers, and that
$$\frac{1}{a + 3} + \frac{1}{b+4} + \frac{1}{c+5} = \frac{7}{12},$$
compute:
$$\frac{a}{a+3} + \frac{b}{b+4} + \fra... | It is given that $a,b,c \geq 1$. If $a \geq 2$. Then:
$$\frac{1}{a+3}+\frac{1}{b+4}+\frac{1}{c+5} \leq \frac{1}{5}+\frac{1}{5}+\frac{1}{6}=\frac{17}{30} < \frac{7}{12}$$
So $a=1$ and
$$\frac{1}{b+4}+\frac{1}{c+5}=\frac{1}{3}$$
If $b \geq 3$, we get:
$$\frac{1}{b+4}+\frac{1}{c+5} \leq \frac{1}{7}+\frac{1}{6} = \frac{13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Minimize $\frac{2}{1-a}+\frac{75}{10-b}$
Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum
value of $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$
WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.
Consider making a transformation like this
$$\dfrac{2}{1-a... | I come up with a solution which is based on the clue I posted above.
\begin{align*}
\frac{2}{1-a}+\frac{75}{10-b}&=\frac{2}{1-a}+\frac{15}{2}\cdot\frac{b}{10-b}+\frac{15}{2}\\
&=\frac{2\cdot \frac{a}{2}\cdot \frac{a}{2}}{(1-a)\cdot \frac{a}{2}\cdot\frac{a}{2}}+\frac{15}{2}\cdot\frac{b\cdot b}{(10-b)\cdot b}+\frac{15}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding an Explicit Formula for Probability Given the Recursive Formula I am attempting to solve a problem asking for the formula to find a probability, $p(n)$, $n$ being 3, 4, 5,... I was able to determine that $p(3) = 1/4$ and the $p(n)$ given not $p(n-1)= \frac{n-2}{2(n-1)}$. From that, I can write the recursive fo... | I will post a solution by @robjohn
Let $t(n) = 1 - p(n)$, then we rewrite the equation as
$$ t(n) = \frac{n}{2(n-1)} t(n-1) = \frac{n}{2(n-1)}\cdot\frac{n-1}{2(n-2)}\cdot t(n-2)$$
you proceed to a generic step $k$
$$t(n) = \frac{n}{2(n-1)}\cdot\frac{n-1}{2(n-2)}\cdot\ldots\cdot\frac{n-k+1}{2(n-k)}\cdot t(n-k) = \frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\int_0^{2\pi} \frac{1}{3 + \cos x} dx$. I have to find the integral
$$\int_0^{2\pi} \dfrac{1}{3 + \cos x} dx$$
I tried using the Weierstrass subtitution, but replacing the bounds, I get:
$$t_1 = \tan \dfrac{0}{2} = \tan 0 = 0$$
$$t_2 = \tan \dfrac{2 \pi}{2} = \tan \pi = 0$$
Resulting in the integral:
$$\int_0^0 \... | By the tangent half-angle substitution we obtain:
\begin{align}2\int_0^{\pi} \dfrac{1}{3 + \cos x} \,dx&=2\int_0^{\infty}\frac{1}{3+\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}\,dt\\&=
2\int_0^{\infty}\frac{1}{t^2+2}\,dt\\&=
2\left(\frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt2{}}\right)\right)\Bigg|_0^{\infty}
\\&=2\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all values of $m$ such that the equation $ mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$ has nonnegative roots.
Find all values of $m$ such that the equation $$\large mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
For an equation to have nonnegative roots, it mustn't only have negative roots.
Let $y... | Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $x^2$
$$0=m(x^2+16/x^2)+x+4/x+8m-1=m(x+4/x)^2+x+4/x-1$$
Now as $x\ge0,$ $$\dfrac{x+4/x}2\ge\sqrt{x\cdot4/x}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of tangents for $x^3+y^3-3xy=0$ at $x=0,y=0$ $$x^3+y^3-3xy=0$$
Find the equation of tangents at $x=0,y=0$
My attempt is as follows:-
Attempt $1$:
$$3x^2+3y^2\dfrac{dy}{dx}-3\left(x\dfrac{dy}{dx}+y\right)=0$$
$$\dfrac{dy}{dx}(y^2-x)=y-x^2$$
$$\dfrac{dy}{dx}=\dfrac{y-x^2}{y^2-x}$$
but when placing $x=0,... | Continue with what you obtained
$$y'= \frac{y-x^2}{y^2-x} $$
and note that the near the origin, $y' = \frac yx$, or
$$\frac{y-x^2}{y^2-x} = \frac yx \implies x^3+y^3 = 2xy$$
Then, substitute $x^3+y^3 - 3xy=0$ to obtain the equation near the origin,
$$xy=0$$
which represents the $x$- and the $y$-axes and also the ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving the sequence $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$: proving that $2+2a_n$ is a perfect square Question:
Let $a_1=a_2=97$ and $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$ for $n>1$. Prove that
(a) $2+2a_n$ is a perfect square, and (b) $2+\sqrt{2+2a_n}$ is a perfect square.
I changed the ... | Simply redoint the argument of @Martin R: so I understand it better.
If $a = \frac{1}{2}(s +1/s)$ then $a^2-1= \left(\frac{1}{2}(s - 1/s)\right)^2$. Now for
$a = \frac{1}{2}(s + 1/s)$, $b = \frac{1}{2}(t + 1/t)$ ( $s$, $t> 1$) we get
$$a b + \sqrt{(a^2-1)(b^2-1)}= \frac{1}{4} (s+1/s)(t+ 1/t) + \frac{1}{4} (s-1/s)(t-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integer solutions to the product of four consecutive integers $ \bullet \textbf{Question} $
The product of four consecutive integers $ x, x + 1, x + 2, x + 3 $ can be written as the product of two consecutive integers, find all integer solutions for $ x $.
$ \bullet \textbf{Rephrasing} $
I decided to name the other int... | We have $$y^2+y=(x^2+3x)(x^2+3x+2)$$ or$$y^2+y+1=(x^2+3x+1)^2$$ or
$$(2y+1)^2+3=(2x^2+6x+2)^2$$ or
$$(2x^2+6x-2y+1)(2x^2+6x+2y+3)=3.$$
Now, solve a number of systems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $xy''-y'-x^3y=0$ I want to solve $xy''-y'-x^3y=0$. My solution:
$y = z\cdot \exp(kx^2)$
$y' = z'\cdot \exp(kx^2) + z\cdot 2kx\exp(kx^2)$
$y'' = z''\cdot \exp(kx^2) + z'\cdot 4kx\exp(kx^2) + z\cdot (2k+4k^2x^2)\exp(kx^2)$
Plug in:
$z''\exp(kx^2)+z'(4kx-\frac{1}{x})\exp(kx^2)+z(2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2)\e... | We can really smoke this with the substitution $u=x^2$:
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=2x\frac{dy}{du}$$
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(2x\frac{dy}{du}\right)=2\frac{dy}{du}+4x^2\frac{d^2y}{du^2}$$
So
$$x\frac{d^2y}{dx^2}-\frac{dy}{dx}-x^3y=4x^3\frac{d^2y}{du^2}-x^3y=4x^3\left(\frac{d^2y}{du^2}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$... | Note that
$$\frac{f(x)}{g(x)}=\frac{(x-1/x)^2+2}{x-1/x}=x-\frac1x+2\frac{1}{x-\frac1x}.
$$
Letting $u=x-1/x$ we take the derivative of $u+2/u$ to get $u'-\frac{2}{u^2}u'=u'(1-2/u^2)=0$. As $u'$ is always positive we must have $2=u^2=(x-1/x)^2$ which is easily solved; $x=\pm\sqrt{2\pm\sqrt3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negati... | Disclaimer: Since you are saying that it is your last attempt on Webwork, so I would suggest that you should read through the steps carefully to see if there is any typo or mistake on my part.The approach is along the same lines however.
Since $4x-x^2=-(x^2-4x+2^2-2^2) =-(x-2)^2+2^2 $
So
$\int \frac{x^2}{\sqrt{4x-x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
An approximation related to Euler's constant and the Harmonic number Let's consider Euler's constant $\gamma$, i.e.,
$$\gamma=\lim_{n\to \infty} \sum_{k=1}^n\frac{1}{k}-\ln(n).$$
Prove the following approximation:
$$\sum_{k=1}^{m-1}\frac{1}{k}-\ln(m)+\frac{1}{2m}+\frac{1}{12m^2}\approx \gamma.$$
The above approx... | We have
\begin{align}
\sum\limits_{k = 1}^m {\frac{1}{k}} - \log m &= \sum\limits_{k = 1}^m {\frac{1}{k}} - \log \prod\limits_{k = 2}^m {\frac{k}{{k - 1}}}
\\
&= \sum\limits_{k = 1}^m {\frac{1}{k}} - \sum\limits_{k = 2}^m {\log \frac{k}{{k - 1}}} \\&= 1 + \sum\limits_{k = 2}^m {\left[ {\frac{1}{k} - \log \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integer solutions of $\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $ (AusPol 1994) Find all integer solutions of
$$\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $$
Attempt:
I noticed that $a,b,c>0$ or $a,b,c<0$ can't happen. Besides, if one of them is zero, we can find some solutions. Suppose $c=0$, we get
$$(a+b)\f... | Rewriting, we have:
$$(a+b)(b+c)(c+a)+2(a+b+c)^3+2abc=2$$
The case of $c=0$ was solved using the fact that the LHS was divisible by $a+b$. Then, we could say that it was divisible by $(a+b+kc)$ since $kc=c=0$. Since the equation is symmetric, we can guess:
$$(a+b)(b+c)(c+a)+2(a+b+c)^3+2abc=m(a+b+kc)(a+kb+c)(ka+b+c)$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$ using geometry I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.
$$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$
My Attempt:
Let ... | Building on @WETutorialSchool's hint, start from the identity$$i\tan\frac{\theta}{2}=\frac{2i\sin\frac{\theta}{2}\exp\frac{i\theta}{2}}{2\cos\frac{\theta}{2}\exp\frac{i\theta}{2}}=\frac{\exp i\theta-1}{\exp i\theta+1}.$$Define $z:=\exp\frac{2i\pi}{2n+1}$ so$$\frac{z^k-1}{z^k+1}=i\tan\frac{k\pi}{2n+1},\,\frac{z^{2n+1-k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
A and B commute. Delete the same row and column from each. Do they still commute? Suppose that two matrices $A$ and $B$ commute. If I delete the $i$th row and $i$th column from each, do they necessarily commute? Does the answer change if we guarantee that $A$ and $B$ are Hermitian?
For example, consider the matrices
... | The answer is no in general. Consider
$$
A=
\begin{bmatrix}
6 & 1 & 8 \\
7 & 5 & 3 \\
2 & 9 & 4
\end{bmatrix}
\quad\text{and}\quad
B=
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix}
.$$
Then $AB$ computes the row sums of $A$ while $BA$ computes the column sums of $A$. Since this $A$ is a magic square ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the locus of $z^2+\bar{z}^2=2$? I have to prove that it's the equation of an equilateral hyperbola
$$z^2+z^{-2}=2$$
I try this
$z^2+z^{-2} +2 = 2+2$
$(z^2+1/z^2 +2 ) = 4$
$(z+1/z)^2=4$
$ z+1/z= 2 $
$ x+yi + 1/(x+yi) = 2 $
$ ((x+yi)^2+1)/(x+yi)=2$
$ x^2+2xyi-y^2+1=2x+2yi$
$ x^2-2x+1 +2xyi -y^2=2yi $
$(x-1)^2 +2... | The locus of $$z^2+(\bar z)^2=2 \implies (x+iy)^2+(x-iy)^2=2 \implies x^2-y^2=1$$ which is a hyperbola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find the div(curl $v$)?
let $v= ( v_1 ,v_2,v_3)$ be a vector field on $\mathbb{R}^3$ where $v_1 = \sqrt {1+x^2+y^2} , v_2= \sqrt {1+z^2} $and $v_3= \sqrt{ 1 +x^2y^2z^2}$. Evaluate div(curl $v$)
My attempt : $$Cur v= \begin{bmatrix} i &j&k\\ \frac{dv_1}{dx}& \frac{dv_2}{dy} & \frac{dv_3}{dz} \\ \sqrt {1+x^2... | It is a general fact that $\operatorname{div}(\operatorname{curl}(F))=0$. You can verify this by expanding. Let $F=(P,Q,R)$. Then
$$\operatorname{curl}(F)=(R_y-Q_z)i-(R_x-P_z)j+(Q_x-P_y)k$$
by definition. Then
$$ \operatorname{div}(\operatorname{curl}(F))=R_{xy}-Q_{xz}-R_{yx}+P_{yz}+Q_{zx}-P_{zy}$$
$$=(R_{xy}-R_{yx}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n + 2)2^{n-1}$. How can I prove the following identity?
$$\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n +
2)2^{n-1}$$
I thought about differentiating this:
$$(1 + x) ^ n = \sum_{k = 0} ^ {n} \binom{n}{k}x^k$$
and then evaluating it at $x =... | Here's an alternative proof that does not depend on derivatives:
\begin{align}
\sum_k (k+1) \binom{n}{k}
&= \sum_k k\binom{n}{k} + \sum_k \binom{n}{k} \\
&= \sum_k n\binom{n-1}{k-1} + \sum_k \binom{n}{k} \\
&= n 2^{n-1} + 2^n \\
&= (n+2) 2^{n-1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How $\sum_1^6\sin x \equiv2\sin\frac{7}{2}(\cos\frac{5}{2}+\cos\frac{3}{2}+\cos\frac{1}{2})$? I just learned this cool trick but I can't figure out why it works, obviously I know how it's done:
1 + 2 + 3 + ... + n ( I know the no of terms have to be even for it to work in pairs)
$\sin$(1)+ $\sin$(2) + $\sin$(3)+... = ... | For the first question, you just need to look at the values of $\alpha$ and $\beta$. As you have noted, they are paired so the sum is always $n+1$. The first few $(\alpha,\beta)$ pairs are $$(n,1)\\(n-1,2)\\(n-2,3)\\\vdots$$ The sums are always $n+1$ and the differences are $$n-1\\n-3\\n-5\\\vdots$$
Symbolically, we... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $xy>x+y$ if $x,y \in (2,\infty)\subseteq \Bbb R$ I am new to proofs. This is my first proof-based mathematics class, and it is a hard transition from my high school classes. I am a second semester freshman.
My original question was:
Prove that there is a real number with the property that for any two
la... | If $x>z$ and $y > z$ then $a=x-z> 0$ and $b=y-z>0$.
And $x+y = (z+a) + (z+b) = 2z +(a+b)$. And $xy = (z+a)(z+b) = z^2 + (a+b)z + ab$.
We want to find a $z$ so that $2z + (a+b) < z^2 +(a+b)z + ab$ for all possible positive $ab$.
This will happen if and only if $0 < (z^2-2z) + (a+b)(z-1) + ab=z(z-2) + (a+b)(z-1) + ab$ f... | {
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Calculus 2: Integration by Parts Stuck on Integral of Product With ArcTan Inside I'm stuck on the following problem:
$$\int_0^{1/3} y \tan^{-1}(3y)\,dy$$
I think my last line in my work below is correct but I don't know what to do beyond that.
A step through of the problem would be appreciated.
| Integrate by parts $\left(f = \tan^{-1}3y, g' = y \Rightarrow f' = \frac{3}{9y^2 + 1}, g = \frac{y^2}{2}\right)$:
$$\int y \tan^{-1}(3y) dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \int \frac{3y^2}{2(9y^2 + 1)} dy$$
$$= \frac{y^2 \tan^{-1}(3y)}{2} - \frac{3}{2} \int \frac{y^2}{9y^2 + 1} dy$$
Write $y^2$ as $\frac{1}{9}(9y^2... | {
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Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$. This is a homework question. I have to find two limits:
i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$
ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac... | The second limit can be squeezed as well with :
$$n^4 \leq \sqrt{n^8+k} \leq n^4+1$$
Thus
$$n\left(\sum_{k=1}^n \frac{k^3}{n^4+1}-\frac{1}{4}\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{n^4}-\frac{1}{4}\right)$$
or
$$n\left[\frac{n^2(n+1)^2}{4(n^4+1)}-\f... | {
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"source": "stackexchange",
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How to prove the asymptotic expansion $\overline{H}_n \sim \log(2) -(-1)^n\left (\frac{1}{2n}-\frac{1}{4 n^2} +\frac{1}{8n^4}\mp\ldots\right)$? It is well-known that the harmonic sum $H_{n}= \sum_{k=1}^{n}\frac{ 1}{k}$ has the following asymptotic expansion for $n\to\infty$
$$H_n = \sum_{k=1}^{n}\frac{1}{k}\sim \gamma... | My idea was to express $\overline{H}_k$ by $H_k$ and then use the asyptotic expansion of $H_k$.
Indeed, $\overline{H}_n$ can be expressed as follows ($m=1,2,3,\ldots$}
$$\overline{H}_{2m} = H_{2m} -H_{m}\tag{5a}$$
$$\overline{H}_{2m+1} = H_{2m+1} -H_{m}\tag{5b}$$
The (simple) proof is left as an exercise to the reader.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$
I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)
I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) ... | Here is a nice solution that utilizes symmetry instead of $u$ susbsitution. Notice that the domain of this function is at most $\left[0,\frac{1}{2}\right]$, so for the moment let's define
$$F(x) \equiv \int_0^x \arcsin\left(\sqrt{\frac{t}{1-t}}\right)\:dt$$
We could do this integral directly, or since the integrand has... | {
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"source": "stackexchange",
"question_score": "3",
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$f(x)=\frac{\sin x}{x}$, prove that $|f^{(n)}(x)|\le \frac{1}{n+1}$ This is a homework question.
Let $f: (0, \infty) \to \mathbb{R}$ with $f(x)=\dfrac{\sin x}{x}$. I
have to prove that
$$|f^{(n)}(x)|\le \frac{1}{n+1}$$
where $f^{(n)}$ is the nth derivative of $f$.
I started to do a few derivatives:
$f^{(1)}(x)=\d... | I will assume (since it's not specified) that $n$ is a non-negative integer.
Claim: For $f:(0,\infty)\to \mathbb{R}$, defined by $f(x)=\dfrac{\sin x}{x}$ and any non-negative integer $n$, we have:
$$f^{(n)}(x)=\frac{1}{x^{n+1}}\int_0^xu^n\cos\left(u+\frac{n\pi}{2}\right)\,du$$
Proof: I will prove this by induction. The... | {
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"source": "stackexchange",
"question_score": "9",
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Why does $3x^4 + 16x^3 + 20x^2 - 9x - 18$ = $(x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3$? $$3x^4 + 16x^3 + 20x^2 - 9x - 18 $$
When simplified I arrive to:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6}) $$
But the math book wrote:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3 $$
with that extra 3 at the end. The graph calculator s... | For any real $k$ we have
$$3x^4+16x^3+20x^2-9x-18=3\left(x^4+\frac{16}{3}x^3+\frac{20}{3}x^2-3x-6\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+k\right)^2-\frac{64}{9}x^2-\frac{16k}{3}x-2kx^2-k^2+\frac{20}{3}x^2-3x-6\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+k\right)^2-\left(\left(2k+\frac{4}{9}\right)x^2+\left(\frac{16k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and... | Hint:
Clearly, $\sin x\ne0$
$$1=4\cos x\cos2x\cos3x$$
$$\implies\sin x=2(\sin2x)\cos2x\cos3x=\sin4x\cos3x$$
$$\implies2\sin x=2\sin4x\cos3x=\sin7x+\sin x$$
$$\implies0=\sin7x-\sin x=2\sin3x\cos4x$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Integral $\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx$ I would like to learn more about this integral:
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx=\frac{1}{128}\Big(16\pi \operatorname{C}-21\zeta(3)-4\pi^2\ln(2)\Big)$$
where $\operatorname{C}$ is the Catalan's constant.
I have tried integration by parts , substitution $y=\cos ... | By the Fourier series of $\ln (\cos x)$
$$
\ln (\cos x)=-\ln 2+\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \cos (2 k x),
$$
we have $$
x \ln (\cos x)=-x \ln 2+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} x \cos (2 k x)
$$
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} x \ln (\cos x) d x&=-\ln 2 \int_{0}^{\frac{\pi}{4}} x d x+\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572727",
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"source": "stackexchange",
"question_score": "2",
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If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ I have the polynomial $x^4+12x-5$ with the roots $x_1,x_2,x_3,x_4$ and I want to find the polynomial whose roots are $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$.
I found the roots $x_1=-1+\sqrt{... | Eliminate $x$ out of the system
$$s^4 + 12 s - 15 = 0 \\
t^4 + 12 t - 15 =0 \\
x- (s+t) = 0$$
and get an equation of degree $10$ in $x$
$$x^{10} + 96 x^7 - 60 x^6 - 144 x^4 + 1920 x^3 - 1600 x^2 - 13824 x + 11520=0 \ \ (*)$$
Now, some of these solutions ($4$ of them) of the above are in fact solution of the equation i... | {
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"question_score": "4",
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"answer_id": 4
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Divisibility by $33^{33}$ Let $P_n=(19+92)(19^2+92^2)\cdots(19^n+92^n)$ for each positive integer $n$. Determine , with proof the least positive integer $n$, if it exists , for which $P_n$ is divisible by $33^{33}$.
I have made no progress concrete enough to show .
| COMMENT:
$19^n+92^n ≡ 8^n+4^n \ mod (11)$
⇒ $4^n(2^n+1)=11 k$
One solution of this equation is $n=5$, $k=3 \times 1024$, we have:
$4^5(2^5+1)=1024\times 33$
Therefor for $n=5$ and all powers such as $5(2t+1)$ , $19^n+92^n ≡ 0 \mod (33)$ due to geometric progression:
$(19^5)^{2t+1}+(92^5)^{2t+1}=(19^5+92^5)[(19^5)^{2t}-... | {
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Prove $ \frac{1}{201} < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac{1}{100}$ Prove: $\displaystyle \frac{1}{201} < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac{1}{100}$
What i was trying to do:
On the right side:
$$
\int^{100}_{0} \frac{e^{-x}}{100+x} dx < \int^{100}_{0} \frac{1}{100+x} dx = ln(100+x) |^{100}_0 =... | Using the estimation as already given in @BrianMoehring's comment and noting that
*
*$\int_0^{100}e^{-x}dx = 1- \frac{1}{e^{100}}$ you have
$$\frac 1{200}\left(1- \frac{1}{e^{100}}\right) < \int^{100}_{0} \frac{e^{-x}}{100+x} dx < \frac 1{100}\left(1- \frac{1}{e^{100}}\right) < \frac 1{100}$$
The lower bound seem... | {
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"source": "stackexchange",
"question_score": "1",
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Remainder when $2^{55}+1$ is divided by $33$ There's this problem I encountered in a math Olympiad for my country.
Find the remainder when $2^{55}+1$ is divided by $33$.
My approach was to consider $2^{55}$ as the sum of numbers in the 56th row of Pascal's triangle. Then I showed that apart from 1 and 55, all other n... | Sticking to Pascal's triangle, you can just start calculating $2^k \pmod{33}$ row by row until you see a 'pattern' and can then get lazy.
$\quad 2^0 \equiv \;\,1 \pmod{33}$
$\quad 2^1 \equiv \;\,2 \pmod{33}$
$\quad 2^2 \equiv \;\,4 \pmod{33}$
$\quad 2^3 \equiv \;\,8 \pmod{33}$
$\quad 2^4 \equiv 16 \pmod{33}$
$\quad 2^5... | {
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"answer_id": 4
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Compute $S_n=\sum_{k=1}^{n} {n \choose k} k^2$ Compute $S_n=\sum_{k=1}^{n} {n \choose k} k^2$.
This is took from Arthur Engel’s book, from the enumerative combinatorics chapter. I can’t understand the author’s explanation. He says the sum represents the number of ways to choose a comittee, a chairman, and its secretary... | Here is the algebric way to prove it :
Let $ n $ be a positive integer greater than $ 1 $, know : $$ \left(\forall k\in\left[\!\left[1,n-1\right]\!\right]\right),\ k^{2}\displaystyle\binom{n-1}{k}=k\left(n-k\right)\displaystyle\binom{n-1}{k-1}$$
Thus, \begin{aligned} \displaystyle\sum_{k=1}^{n-1}{k^{2}\displaystyle\bin... | {
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Finding xy+yz+zx such that the given determinant = 0
$x≠y≠z$
$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix} = 0$
Then xy+yz+zx = | A. x+y+z | B. $xyz$ | C. $xyz\over(x+y+z)$ | D. $(x+y+z)\over xyz$ |
Given Ans - D
What I did first was R1->R1-R3 & R2->R2-R3 and throwing (x-z) and (y-z) to the 0... | We can take advantage of this being a multiple choice question.
Expanding out the determinant $$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}$$ will give products that look like $xy^3(z^4-1)$, which give us six degree-$8$ terms (such as $xy^3z^4$) and six degree-$4$ terms (such as $-xy^3$).
We know... | {
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"source": "stackexchange",
"question_score": "4",
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Prove that in acute triangle : $\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$ Im going to prove this identity in acute triangle :
Let $ABC$ acute triangle , $A,B,C$ are angles then :
$$\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$$
I know that $A,B,C<\frac{π}{2}$
$$\sin A\cos (B-C)=\sin A(\cos B\cos C+\sin A\s... | First, we need to derive this little identity:
$\sin (x +y) = \sin x\cos y - \cos x\sin y\\
\sin (x - y) = \sin x\cos y + \cos x\sin y\\
\sin (x +y) + \sin(x-y) = 2\sin x\cos y\\
\sin x\cos y = \frac{\sin (x +y) + \sin(x-y)}{2}$
Now we can apply it with $x = A, y = B-C$
And while we are at it, we can derive similar ide... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{... | You can work out a continued fraction form for $2A$ and compare it with $B$.
Start with
$A=1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}...}}}$
Then we multiply by 2, by multiplying the initial $1$ by $2$ and dividing the denominator by $2$ at the first layer:
$2A=2+\dfrac{1}{(1/2)(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}... | {
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how to calculate $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos^2x}$? $$
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{1}{1+\cos ^2x}\rm{dx}+\int_{\frac{\pi }{2}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}}
$$
I want to split this integral,$
\i... | A possible, maybe more convenient way is as follows:
*
*First shift the integral using $x=\frac{\pi}{2} + u$:
$$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos^2x}= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{du}{1+\sin^2u}= 2\int_{0}^{\frac{\pi}{4}}\frac{du}{1+\sin^2u}$$
*Now, use $t=\tan u$ and note that $\... | {
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Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\righ... | Your idea is good; you want to find all positive integers $k$ for which there is precisely on integer between the roots of
$$5x^2-2kx+1=0.$$
Then the distance between the roots can be at most $2$, where the distance between the roots is precisely
$$\frac{1}{5}\sqrt{(-2k)^2-4\cdot1\cdot5}=\frac25\sqrt{k^2-5},$$
as you a... | {
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"url": "https://math.stackexchange.com/questions/3588314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability Two People are Born on the Same Date (alternative way) What is the probability two people (individuals) will have the same exact birthday?
There are 365 days in a year and I assume that any person can be born on any random day, so uniformly.
I like to use a slots method when I look at combinations/permutat... |
$\{ \text{Slot 1 + Slot 2} \} = \frac{ \quad 0 \quad }{1} \frac{
\quad 1 \quad }{2} \frac{ \quad 1 \quad }{3} \quad \cdot \cdot
\cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{
\quad 0 \quad }{365} $
There $365 \choose 2$ ways of arranging two $1$'s and three hundred
sixty three $0... | {
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"url": "https://math.stackexchange.com/questions/3589580",
"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+... | From your third step you get $c=0$ and $a=1/2, \implies 1/2+b+0=3 \implies b=5/2$.
Next, $f(n)=n^2/2+5n/2$, then
$$S_m=\sum_{n=1}^{m} (n^2/2+5n/2)= \frac{m(m+1)(2m+1)}{12}+\frac{5m(m+1)}{4}$$
$$\implies S_m=\frac{m(m+1)(m+8)}{6}$$
then $$S_{10}=10.11.18/6=330$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}... | If you are allowed to integrate,
then, for $a > 0$,
$\begin{array}\\
\int_n^{n+1} x^{-a} dx
&= \dfrac{x^{-a+1}}{-a+1}|_n^{n+1}\\
&= \dfrac{(n+1)^{-a+1}-n^{-a+1}}{-a+1}\\
\text{and}\\
\int_n^{n+1} x^{-a} dx
&\lt n^{-a}
\qquad\text{since } x^{-a} \text{ is decreasing}\\
\end{array}
$
Putting $a = \frac13$.
this becomes
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
3 circles of radii 1,2 and 3 and centers A, B and C respectively, touch each other.......
3 circles of radii 1,2 and 3 and centers A, B and C respectively, touch each other. Another circle whose center is P touches all 3 externally and has radius $r$. Also $\angle PAB=\theta$ and $\angle PAC=\alpha$. Prove that
$\cos ... | It should be $$\cos\theta=\frac{9+(1+r)^2-(2+r)^2}{2\cdot3\cdot(1+r)}=\frac{6-2r}{6(1+r)}=\frac{3-r}{3(1+r)}.$$
Also, $$\cos\alpha=\frac{16+(1+r)^2-(3+r)^2}{2\cdot4\cdot(1+r)}=\frac{2-r}{2(1+r)}$$ and since $$\alpha+\theta=90^{\circ},$$ we obtain:
$$\left(\frac{3-r}{3(1+r)}\right)^2+\left(\frac{2-r}{2(1+r)}\right)^2=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For acute $\triangle ABC$, prove $(\cos A+\cos B)^2+(\cos A+\cos C)^2+(\cos B+\cos C)^2\leq3$
Prove that, in an acute $\triangle ABC$,
$$(\cos A+\cos B)^2+(\cos A+\cos C)^2+(\cos B+\cos C)^2\leq3$$
I tried this, but I can't to this.
I used $AM\geq GM$ and got $$3\geq\cos(A-B)+\cos(A-C)+\cos(B-C)$$
But I can't see... | Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.
Thus, we need to prove that
$$\sum_{cyc}\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}\right)^2\leq3$$ or
$$\sum_{cyc}\left(\frac{x}{2\cdot\sqrt{\frac{x+z}{2}}\cdot\sqrt{\frac{x+y}{2}}}+\frac{y}{2\cdot\sqrt{\frac{y+z}{2}}\cdot\sqrt{\frac{x+y}{2}}}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$ 5.4
Can somebody verify this solution for me?
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$
The area under the graph of $f(x)$ between $x=-2$ and $x=2$ is exactly equal to:
$\int_{-2}^2 (1-2x... | $$
\int_{-2}^{2}{(1-2x)^{2}dx}\\=\int_{-2}^{2}{1-4x+4x^2dx}\\=\int_{-2}^{2}{1dx}+\int_{-2}^{2}{4xdx}+\int_{-2}^{2}{4x^2dx}
$$
Now lets solve those integrals.
$$
\int_{-2}^{2}{1dx}=4\\
$$
since it has only a constant
$$
\int_{-2}^{2}{4xdx}=0\\
$$
since it's an even function.
$$
\int_{-2}^{2}{4x^2}\\
=4\times\int_{-2}^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600218",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given that $a+b+c=0$, show that $2(a^4+b^4+c^4)$ is a perfect square Given that $a+b+c=0$. Show that: $2(a^4+b^4+c^4)$ is a perfect square
MY ATTEMPTS:
I found that when $a+b+c=0$, $a^3+b^3+c^3=3abc$
So I did:
$(a^3+b^3+c^3)(a+b+c)$ -- $a^4+b^4+c^4=-(a^3c+a^3b+ab^3+b^3c+c^3a+c^3b)$
And then I tried to substitute $a^4+b... | Square $a+b+c=0$
\begin{eqnarray*}
a^2+b^2+c^2=-2(ab+bc+ca).
\end{eqnarray*}
Square this
\begin{eqnarray*}
a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=4(a^2b^2+b^2c^2+c^2a^2)+8abc(a+b+c)
\end{eqnarray*}
The last term is zero ... rearrange
\begin{eqnarray*}
2(a^4+b^4+c^4)=4(a^2b^2+b^2c^2+c^2a^2)=(a^2+b^2+c^2)^2.
\end{eqnarray*... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
| Suppose $2k^3<3^k$. Then
\begin{align}
2(k+1)^3&=2(k^3+3k^2+3k+1)\\
&=2k^3+6k^2+6k+2\\
&<3^k+6k^2+6k+2\\
&< 3^k+k^3+k^2+k\\
&<3^k+4k^3\\
&<3^k+2\cdot3^k\\
&=3^{k+1}
\end{align}
Note that we use in the middle that $6\le k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Minimizing cubic over elliptic disk without using derivatives Given $x, y$ satisfy $(x-4)^2 + (y-4)^2 + 2xy \leq 32$. Find the minimum value of:
$$P = x^3 + y^3 +3(xy-1)(x+y-2)$$
My attempt:
From $(x-4)^2 + (y-4)^2 + 2xy \leq 32$, I get:
$$(x+y)^2 - 8(x+y) \leq 0$$
$$\implies 0 \leq x+y \leq 8$$
From $P = x^3 + y^3 +3(... | Yes, we can.
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Thus, the condition gives $$(x+y)^2-8(x+y)\leq0$$ or
$$0\leq u\leq4.$$
Now, since $u^2\geq v^2$ it's $(x-y)^2\geq0$, we obtain: $$P=8u^3-6uv^2+3(v^2-1)(2u-2)=8u^3-6v^2-6u+6\geq$$
$$\geq8u^3-6u^2-6u+6=\left(2u-\frac{1}{2}\right)^3-\frac{3}{2}u+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants How can we evaluate
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$
Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by elementary method, so I wo... | Beta+IBP $3$ times+log factorization yields
*
*$S=\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}=\int_0^1 \frac2x \text{Li}_4\left(\frac{x^2(1-x)}2\right) dx\\ =\int_0^1 -\frac{2\left((3 x-2) \text{Li}_3\left(\frac{1}{2} (1-x) x^2\right)\right) \log (x)}{(x-1) x} dx\\ =\int_0^1 \frac{2\left((3 x-2) \text{Li}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Finding number of integral solutions to an equation.
Find the number of integral solutions to:
$$x^2+y^2-6x-8y=0.$$
My attempt:
The equation can be rewritten as:
$$x^2+y^2-6x-8y+9+16=25,$$
basically adding 25 to both sides, or equivalently,
$$(x-3)^2+(y-4)^2=25.$$
This is a Pythagorean triplet. The only triplet of ... | In the non-negatives,
$$n^2+m^2=25$$ is indeed solved by
$$(0,5),(3,4),(4,3),(5,0),$$ as you can check by exhaustive search on $n$.
Then the signed solutions are
$$(0,\pm5),(\pm3,\pm4),(\pm4,\pm3),(\pm5,0).$$
There are $2+2^2+2^2+2$ of them, that you shift by $(3,4)$ to get $(x,y)$.
You can even minimize the "effort" ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. Your answer should no longer include any logarithms.
I noted that $\log_5 10=\frac{1}{\log_{10} 5}.$
I also noted that $\log_{5} 10=\log_5 2+\log_5 5=... | $$ \log_{10} 5 = \log_{8} 5 \times \log_{10}{8}$$
$$\log_{10}{8}=\frac{1}{\log_{8}{10}} = \frac{1}{\log_{8} 2+\log_8 5}$$
$$\log_8 5 = \log_8 3\times \log_3 5 = PQ, \log_{8} 2=\frac{1}{3}$$
$$ \log_{10} 5 = PQ\times \frac{1}{\frac{1}{3}+PQ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Application of lagrange mean value theorem Let $f$ be continuous on $[a,b]$ , differentiable on ($a,b$), and its given $f(a)=a, f(b)=b$. We need to prove the following:
*
*$f'(c1)+f'(c2)=2,$ , for some $c1,c2 \in (a,b)$, and $c1\neq c2$.
*$\frac{1}{f'(c1)}+ \frac{1}{f'(c2)}=2,$ , for some $c1,c2 \in (a,b)$, and $... | For $2$, since $f(a) = a$ and $f(b) = b$, with $a \lt b$, since $f$ is continuous, note the intermediate value theorem says there's a point, call it $a \lt d \lt b$, where
$$f(d) = \frac{a + b}{2} \tag{1}\label{eq1A}$$
Now, use the Lagrange mean value theorem on the sub-intervals $(a, d)$ and $(d, b)$ to get for that s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\fra... | I think the answer is not equal to $16$.
Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3u$, $\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=3v^2$ and $\frac{1}{abc}=w^3$.
Thus, $u=\frac{1}{3},$ $v^2=-\frac{1}{3}$ and $w^3=2$.
Thus, $$\prod_{cyc}\left(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\right)=\prod_{cyc}\left(\frac{1}{a^3}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find the minimum value of $a^2+b^2+c^2+2abc$ when $a+b+c=3$ and $a,b,c\geq0$.
Given $a,b,c\geq0$ such that $a+b+c=3$, find the minimum value of $$P=a^2+b^2+c^2+2abc.$$
It seems like the minimum value of $P$ is $5$ when $a=b=c=1$, but I can find at least one example where $P<5$.
My attempt:
Without loss of generality... | Your attempt already contains very good observations, and is quite close to a complete solution. All that remains is an analysis of when both inequalities are equalities. You use the following two inequalities:
\begin{eqnarray*}
a^2+b^2+c^2+2abc&\geq&a^2+b^2+c^2+2bc\tag{1}\\
a^2+(b+c)^2&\geq&\frac{(a+b+c)^2}{2}\tag{2}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+... | The common method is to complete:
$$\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}\cdot \frac{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}\cdot \frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} + \sqrt{2}}=\\
\lim\limits_{x \to2} \frac{\sqrt{x} + \sqrt{2}}{\sqrt[3]{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges Show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges.
I was able to show $\int_{2}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges, comparing it with the function $\frac{1}{x^{3/2}}$. I have trouble showing that $\int_{1}^{2} \frac{1}{\sqrt{x^4-x}} dx$... | The problem seeming to be at $x=1$, make the Taylor series
$$\frac{1}{\sqrt{x^{4} - x}} =\frac{1}{\sqrt{3} \sqrt{x-1}}-\frac{\sqrt{x-1}}{\sqrt{3}}+\frac{5 (x-1)^{3/2}}{6
\sqrt{3}}-\frac{2 (x-1)^{5/2}}{3 \sqrt{3}}+\frac{13 (x-1)^{7/2}}{24
\sqrt{3}}+O\left((x-1)^{9/2}\right)$$ Integrate termwise to get
$$\frac{2 \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Computing the Fourier transform of a complex-valued function I have the following function: $e^{-(a+bi)|x|^2}$.
While trying to compute the fourier transform of the following function, I know that fourier transform of the real part remains the same and the result would be $\sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a}$. I'm not... | Completing the square in the exponent and the well known integral of $e^{-\pi x^2}$ are tools you can use:
$\begin{align}{F}\left\{e^{-(a+bi)|x|^2}\right\} &= {F}\left\{e^{-ax^2}e^{-ibx^2}\right\}\\
\\
&= \dfrac{1}{2\pi}{F} \left\{e^{-ax^2}\right\} * {F}\left\{e^{-ibx^2}\right\}\\
\\
&= \dfrac{1}{2\pi}\left( \sqrt{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629803",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x... | Let $g(x)=f(x-3)$. Then $g(x)=x^{\frac13}(x+30)$
The Taylor Series of $x^\frac13$ is
$$\begin{split}x^{\frac13}&=a^{\frac13}+\sum^\infty_{n=1}\frac{a^{\frac13-n}\prod^n_{m=1}\big(\frac43-n\big)}{n!}(x-a)^n\\&=a^\frac13+\frac13a^{-\frac23}(x-a)+...+\frac{a^{-\frac{236}3}\prod^{79}_{n=1}\big(\frac43-n\big)}{79!}(x-a)^{79... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Finding $\lim_{n\to\infty}\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)$ I have a trouble with this limit of the infinite product:
$$\lim _{n \to\infty}\left(1-\frac{1}{1 \cdot 2}\right)\left(1-\frac{1}{2 \cdot 3}\right) \cdots\left(1-\frac{1}{n(n+1)}\right)$$
My attempt:
We have
$$\prod_{n=1}^{\infty}\left(1-\f... | I continue from where you left. The infinite product representation of the sine function can be used to finish the calculation:
\begin{align*}
& \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{ - 1 - a_2 }}{{n + 1}}} \right)} = \mathop {\lim }\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying ... | Fix $a,b > 0$, and let
$$
f(x)
=
\frac{ab\,\sin x}{2\sqrt{(a^2+b^2+2ab\cos x)(a+b-\sqrt{a^2+b^2+2ab\cos x})}}
$$
As $x$ approaches zero from the right, we have $f(x) > 0$, so the limit from the right, if it exists, $L$ say, must be nonnegative, in which case, since $f$ is an odd function, the limit from the left will b... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Technique for solving $\frac{ax+b}{cx+d}=\frac{px+q}{rx+s}$ where the sum of numerators equals the sum of denominators I was looking up some shortcuts to solve quadratic equations. I got a technique that applies when the sum of the numerators and denominators are equal, but I am unable to understand the reasoning behin... | Componendo and dividendo (Brilliant) is another method.
Using the third rule with $k=1$, we have:
$$\frac{3x+4+(6x+7)}{3x+4-(6x+7)} = \frac{5x+6+(2x+3)}{5x+6-(2x+3)}$$
$$\Rightarrow \frac{9x+11}{-3x-3} = \frac{7x+9}{3x-3}$$
$$\Rightarrow -9x-11 = 7x+9$$
$$\Rightarrow x = -\frac{5}{4}$$
which is true in general, when we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $ Question -
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$... | Elaborating on the above solution:
We first wish to prove that the last inequality holds. Schur's inequality in its simplest form states that: For non-negative real numbers $x,y,z$, we have $x^3+y^3+z^3+3xyz \geq xy(x+y)+xz(x+z)+yz(y+z)$.
Now, expanding the R.H.S., we obtain: $(X+Y+Z)^3+9XYZ$
$=X^3+3X^2Y+3X^2Z+3XY^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Dot product with a scalar function Example 5: Find $\nabla\cdot\left(\frac{\boldsymbol r}{r^2}\right)$ where $\boldsymbol r(x,y,z)=x\boldsymbol i+y\boldsymbol j+z\boldsymbol k$, $r=\sqrt{x^2+y^2+z^2}$.
In this case, why can we not bring the scalar function $r^{-2}$ outside the dot product?
When I leave the function in... | $$\nabla \cdot \left( \frac{\vec{r}}{r^2} \right) = \nabla \cdot \left( \frac{x}{x^2+y^2+z^2}\vec{i} + \frac{y}{x^2+y^2+z^2}\vec{j} + \frac{z}{x^2+y^2+z^2}\vec{k} \right) \\ = \frac{y^2+z^2-x^2}{(x^2+y^2+z^2)^2} + \frac{x^2+z^2-y^2}{(x^2+y^2+z^2)^2} + \frac{x^2+y^2-z^2}{(x^2+y^2+z^2)^2} = \frac{x^2+y^2+z^2}{(x^2+y^2+z^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find Laurent-series? I am trying to find the Laurent series for the function $\frac{1}{ z (2i - z)}$.
I already obtained for...
(1) ... $2 < | z |$:
$\frac{1}{ z (2i - z)} = \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i - z} \right)
= \frac{1}{2 i} \left( \frac{1}{z} - \frac{1}{z} \cdot \frac{1}{1 - \frac{2i... | Don't worry about the $i$, just treat it as you would any other complex constant. Both series you wrote look correct, but generally it is more clear to write them in the following forms.
For $|z|>2$ we can write:
$$
\frac{1}{z(2i-z)}=-\frac{1}{z^2}\left(\frac{1}{1-\frac{2i}{z}}\right)=-\frac{1}{z^2}\sum_{n=0}^{\infty}\... | {
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Solve $(y-u)u_x + (u-x)u_y = x-y$
Solve
$$(y-u)u_x + (u-x)u_y = x-y, \qquad u=0 \text{ when } xy=1.$$
I tried to solve the equation above using characteristic method
$$\begin{cases}
x'=y-u \\
y' = u - x \\
u' = x-y
\end{cases}. \tag{1}$$
Adding together the first and the second equation in $(1)$... | $$(y-u)u_x+(u-x)u_y=x-y$$
System of characteristic ODEs (Charpit-Lagrange) :
$$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}$$
A first characteristic equation comes from :
$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}=\frac{dx+dy+du}{(y-u)+(u-x)+(x-y)}=\frac{dx+dy+du}{0} \quad\implies\quad dx+dy+du=0$
$$u+x+y=c_1$$
A se... | {
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\righ... | We use the standard pqr method.
Let $p = a + b + c, q = ab+bc+ca, r = abc$. The inequality becomes
$$\left(\frac{pq}{r} - 3\right)^2 \ge 4q \left(\left(\frac{q}{r}\right)^2 - 2\frac{p}{r}\right)$$
or (after clearing the denominators)
$$p^2q^2 + 2pqr - 4q^3 + 9r^2 \ge 0.$$
We split into two cases:
1) $p^2 \ge 4q$: Since... | {
"language": "en",
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Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$
Let $x, y \in \mathbb R$ s. t . $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2-\sqrt{32}$
I know this is a duplicate of another question, but that question has solutions involving calculus and geometry,... | Let $ z=x+iy$. Then, we have $ |z-(1+i)| =2$ from $x^2+y^2=2x-2y+2$, and
$$|z|= |z-(1+i)+(1+i )| \le | z-(1+i)|+ |1+i |= 2+ \sqrt 2=|z|_{max}$$
Thus, the largest possible value is
$$x^2+y^2= |z|_{max}^2 = (2+\sqrt 2)^2=6+4\sqrt2$$
| {
"language": "en",
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Prove using Jensen's inequality that if $abcd=1$ then $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1$ Question -
Let $a, b, c, d$ be positive real numbers such that abcd $=1 .$ Prove that
$$
\begin{array}{c}
\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1... | We can use the Vasc's RCF Theorem. It's like Jensen, but it's not Jensen.
Also, since $f(x)=\frac{1}{(1+e^x)^2}$ has an unique inflection point, we can use Jensen with Karamata, but it's not so nice solution. I am ready to show, if you want.
Indeed, $$f''(x)=\frac{4e^x\left(e^x-\frac{1}{2}\right)}{(1+e^x)^4}.$$
Thus, $... | {
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Show $|a\sqrt{b}-\sqrt{c}|$ equal to zero or larger than $\frac{1}{2}10^{-3}$ when $a, b$ and $c$ are natural numbers strictly less than 100 I need to show that $|a\sqrt{b}-\sqrt{c}|$ is equal to zero or larger than $\frac{1}{2}10^{-3}$ when $a, b$ and $c$ are natural numbers strictly less than 100.
I see why it can b... | If $a\sqrt{b} - \sqrt{c} \neq 0$, then $|a^2b - c| \geq 1$. We see that:
$$
a\sqrt{b} + \sqrt{c} < 100(10) + 10 = 1010
$$
Therefore:
$$
|a\sqrt{b} - \sqrt{c}| = \frac{|a^2b - c|}{a\sqrt{b} + \sqrt{c}} > \frac{1}{1010} > \frac{1}{2000}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$f(t) = \min\{1,t\}$ not operator monotone I want to show that the function on $\mathbb{R}^+$ $f(t) = \min \{1,t\}$ is not operator monotone on the complex $2\times 2$ matrices. My plan is to find matrices $A$ and $B$, $B\geq A$, such that the spectrum of $A$ is in $[0,\infty]$ and the spectrum of $B$ contains elements... | You are on the right track. Actually any pair of matrices such that
*
*$A$ has spectrum on $[0, 1]$,
*$B$ has an eigenvalue on $(1, \infty)$, and
*$B - A$ is of rank $1$
*$A$ and $B$ do not commute
will work as an example.
Indeed, $f(t) \leq t$ together with 2. imply that $B \neq f(B) \leq B$. Since also $f(A) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $\sum\limits_{n=2}^{\infty} \ln \left[1+\frac{(-1)^{n}}{n^{p}}\right](p>0)$ convergent? Is $S=\sum\limits_{n=2}^{\infty} \ln \left[1+\dfrac{(-1)^{n}}{n^{p}}\right](p>0)$ convergent?
I can prove the case when $p\geq 1$: If $p>1$, $\sum\limits_{n=2}^{\infty} \left|\ln \left[1+\dfrac{(-1)^{n}}{n^{p}}\right]\right|\leq ... | 1) If $\frac{1}{2} < p$, the series $\sum_{n=2}^\infty \ln (1 + \tfrac{(-1)^n}{n^p})$ is convergent.
It is easy to prove that $0 \le x - \ln (1 + x) \le 2x^2$ for $x > -\frac{3}{4}$.
Note that $\frac{(-1)^n}{n^p} \ge - \frac{1}{3^p} > - \frac{3}{4}$ for $n\ge 2$.
Thus, we have, for $n\ge 2$,
$$0 \le \frac{(-1)^n}{n^p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving this trigonometric equation I want to solve this equation
$$
\arccos(x)+\arcsin(x^2-x+1)=\pi/2
$$
I write this: for all $x\in [-1,1]$
$\arcsin(x^2-x+1)=\pi/2-\arccos(x)$ then $x^2-x+1=\sin(\pi/2-\arccos(x))=\cos(\arccos(x))=x$
$x^2-x+1=x\Rightarrow x^2-2x+1=0\Rightarrow (x-1)^2=0\Rightarrow x=1 $
Is it true ?... | Yes true indeed, in fact doubly true ( double root) at $ x=1$
Compare terms of two equations with identity
$$ \arccos(x)+\arcsin(x^2-x+1)=\pi/2 $$
$$ \arccos(x)+\arcsin(x)=\pi/2 $$
so we must have
$$ x^2-x+1 = x,\, x^2-2x+1 = 0,\,(x-1)^2=0,\,x= (1,1)$$
Solution can be verified by their graphs at $(x=1)$ their average... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$ For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$
NguyenHuyen gave the following expression$:$
$$\sum \frac12\, \left( 8... | Also, we can use SOS after the following C-S:
$$\sum_{cyc}\frac{a^2}{(b+c)^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}$$ and it's remains to prove that
$$\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}+\frac{1}{4}\geq\frac{a^2+b^2+c^2}{ab+ac+bc},$$ which is sixth degree already and it's obvious by $uv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a function $f$ such that $f(3n)=1$, $f(3n+1)=5/3$, $f(3n+2)=8/3$, where $n$ is an integer
Determine a function $f(x)$ such that :
*
*$f(3n)=1$
*$f(3n+1)=5/3$
*$f(3n+2)=8/3$
where $n$ is an integer
How can you determine it? Thanks for the help!
| I guess, topicstarter wants some closed form description of his function like
$$f(k)=\frac{9}{2}(1*(k+1-3[\frac{k+1}{3}])(k+2-3[\frac{k+2}{3}])+\frac{5}{3}*(k-3[\frac{k}{3}])(k+1-3[\frac{k+1}{3}])+\frac{8}{3}*(k+2-3[\frac{k+2}{3}])(k-3[\frac{k}{3}]).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometric series in proof of Stirling's Formula I am working through a proof of Stirling's Formula in Feller's An Introduction to Probability Theory and it's Applications and am stuck at equation 9.10, where he make a comparison with a geometric series. For full context, he states:
And using the expansion we get:
$$... | Your formula for the sum of a geometric series is slightly off. First of all your formula would be the sum of a geometric series with ratio $(2n+1)^{-1}$, not $(2n+1)^{-2}$. More importantly however, the expression $\frac1{1-r}$ is the sum of $1+r+r^2+...$, however in the current case we're lacking the first term, we i... | {
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"url": "https://math.stackexchange.com/questions/3671295",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.